AS Physics P1 MCQs Preparation Thread.

[abruzzi]

Someone please pleaseee help me with qn no. 25 of the following paper:
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf

Waiting for a reply..

 

[studen12345]

Someone please pleaseee help me with qn no. 25 of the following paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf

Waiting for a reply..

The answer is C
because it is open from one side and closed from other so the frequency would take an even pattern li F0 2F0 4F0 so the lowest frequency is Fo and to get second and third we will use 2F0 and 4Fo and so on an even pattern...

 

[studen12345]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_13.pdf
21 17 30
plzz reply

 

[itallion stallion]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf plz solve q9,q15.thanks alot!!!

 

[hellangel1]

22. Stress=F/A. F and A is given. E=stress/strain. strain=stress/E.
26. Use dsin90=nxlambda and multiply that times 2 (the other 90 degrees) and add 1 (the middle maximum).

Thanks..

 

[hellangel1]

Q22- E=F*l/A*extension
A= (2.5*10^-4)^2 * pi
then substituting all values find l/extension. then find its reciprocal. then *100 to get %.
Q26- d=1/N. N=300000 per m
1/N=n*lambda
n will be 7 then times 2 u get 15

Thanks

 

[Zsiddiqui]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf
need help in q8, q15, q26

 

[SYED ALI ARIZ]

ANY1 WITH GUD PRECISE PHYSICS AND CHEMISTRY NOTES???? PLS POST THEM

 

[Raghad Dia]

Can someone explain for me October 2009 varyant 11 # 12, 14, 15, 22, 26 and 28. I know i'm asking for a lot of questions but i really need help x_x thank you in advance.

 

[h4rriet]

I thought it's tangent to the curve? Can you tell me why you start them all from the origin? I thought answer was B for that reason

I'm afraid I don't know why it's through the origin...

 

[h4rriet]

Thank you for your reply..

I haven't understood qn no. 13, could you please show me the working?

For qn no. 31, I found the rate of flow as t = Q/I = (1.6 x 10^-19) / 4.8 = 3.3 x 10^-20. That is not correct right?

13. For the 5 N, the perpendicular distance from the pivot is 2 metres. For the 15 N, it's 3 metres and for the 10 N, it's 2 m.
In 31, you have to find the rate of flow of ELECTRONS, not charge. So Q=It will turn into Ne=It (N being the number of electrons and e being the charge on one electron).

 

[h4rriet]

I am having problem in november 09/12 can anyone help ,e in these questions q8,9,12,13,14,15

8. In an elastic collision, the relative speed of approach = the relative speed of separation.
9. Use F=m(v-u)/t.
12. All you gotta do is locate distances and determine forces.
13. Vertical component of the velocity of a body in projectile motion is 0 at the highest point, so the K.E. can come only from the horizontal component, which is constant throughout.
14. K.E. of the trolleys when they're moving = the E.P.E. stored in the string initially.
15. Power output/Power input=Efficiency. You have Po and Efficiency; Pi will be x.

 

[h4rriet]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf
21 17 30

21. K.E. doesn't increases because the question states that the rate of flow is constant. E.P.E. increases because pressure increases with depth.
17.
30. The electric field strength between two parallel plates is constant.

 

[itallion stallion]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf plz help with q12 .thanks

 

[abruzzi]

Someone please pleaseee help me with qn no. 10 and 20 of the following paper:
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_1.pdf

Waiting for a reply..

Thanks!

 

[sagar65265]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf plz help with q12 .thanks

If you write down the equations of motion for the two objects (barrel coming down and man going up), you get:

Barrel:

Tension = 120g - 120 * acceleration
Tension = 1200 - 120a

Man:

Tension = 80g + 80a
Tension = 800 + 80a

Equating the two,

1200 - 120a = 800 + 80a
400 = 200a
a = 2 ms^-2.

Since the two objects both start from rest and have the same accelerations (they are connected by the string, so if one moves a distance x, so does the other, in the same time interval), the mans head will be in line with the bottom of the barrel after travelling 9 meters. Therefore,

u = 0 ms^-1 , a = 2 ms^-2 , s = 9m:

v^2 = 0 + 2 * 2 * 9
v^2 = 36
v = 6 ms^-1 = A

Hope this helped!
Good Luck for all your exams!

 

[sagar65265]

Someone please pleaseee help me with qn no. 10 and 20 of the following paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf

Waiting for a reply..

Thanks!

Q. 10 :

If a net force acts on an object, it will cause a change in momentum of the object (usually the change in is in velocity, either in terms of direction or magnitude). The average force between any two times is the total change in momentum that occurs between those two times divided by the time taken for the change in momentum (just like the average speed between two points is the total distance divided by the time taken to travel that distance). In general, the force at any one time is the gradient of the momentum-time graph at that point, but since the average force is being asked for, it's just the gradient of the line joining those two points, in other words, it is B, since that shows the change in momentum divided by the time taken.

Q. 20

Hmmmm. How to explain this? Okay, imagine that the beam actually bends under the stress applied by the hanging load. In that case, can you see that the top of the rod stretches and the bottom of the rod decreases in length - hopefully the diagram attached will help, so this basically means that both X and Y are going to be pulled on, towards the load, while Z will be pushed backwards, i.e. compression - sorry, but I'm not sure how else to explain this!

Hope this helped!
Good Luck for all your exams!

 

[magnesium]

http://www.freeexampapers.com/#A Level/Physics/CIE/2002 Jun Q35 ,11,,21 PLZ SUMONE HELP!

 

[magnesium]

http://www.freeexampapers.com /#A Level/Physics/CIE/2002 Nov Q14?

 

[sagar65265]

http://www.freeexampapers.com /#A Level/Physics/CIE/2002 Nov Q14?

It's basically just balancing moments; Since the entire set up can be assumed to be stationary once the 50g mass is added, the tension in the rope is

Tension = (20/1000) * 9.81 = 0.1962 N

This means that the moment caused by the suspended mass is

Moment = Tension * Distance from pivot = 0.1962 * 0.6 = 0.11772 Nm^-1

This is an Anti-Clockwise moment because the force will pull the left side upwards (and thus in the ACW direction) if it is allowed to act alone.
The weight of the ruler can be assumed to act from its center (at the 50 cm mark), and is a Clockwise moment, because the force will pull the right side of the ruler down (and thus CW) if it is allowed to act alone.
The value of this moment is :

0.1 * 9.81 * 0.1 = 0.0981 Nm^-1

So if the system has to be in equilibrium, then the net moment has to be zero. But the ACW moment due to the 20g mass is more than the CW moment due to the mass of the ruler, so we need to position the 50g mass in such a place that it will exert a CW moment to make the CW moment equal to the ACW moment. This means that the 50g mass will have to be positioned to the right of the pivot, and it will have to exert a moment of magnitude

0.11772 - 0.981 = 0.01962 Nm^-1 (this is the difference in moments in the current situation).

Therefore,

(50/1000) * 9.81 * D = 0.01962
D = 0.04 m = 4 cm to the right of the pivot.

Since the pivot is at 40 cm, the mass will be placed at the 44 cm mark.

Hope this helped!
Good Luck for all your exams!